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In my experience, people usually use "smooth" to mean "as smooth as I need for the upcoming proofs." Those who want to be more formal might insist on smooth meaning $C^\infty$.

While the operator taking $f$ to its Taylor series at some point in its domain uses information about all the partial derivatives of $f$ at once, I don't think I know an example to settle the following question:

Is there anything I can do simultaneously with infinitely many derivatives of a $C^\infty$, not necessarily analytic function?

I hope that's not phrased too vaguely. Since there are $C^\infty$ functions with zero Taylor series, for instance, something like "sure-you can put all its derivatives into an infinite series!" is probably not relevant. I won't be surprised if answer is "No, $C^\infty$ doesn't mean materially more than $C^k$ for sufficiently large $k$." But I'd be very interested to hear otherwise.

Kevin Carlson
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    A nonzero $C^\infty$ function can be compactly supported. Not so for a nonzero analytic function. – Stefan Smith Oct 23 '13 at 22:53
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    Indeed. I'm aware of facts that separate $C^\infty$ from analytic functions-I'm looking more for the other direction, differences between $C^\infty$ and $C^k$. – Kevin Carlson Oct 23 '13 at 23:09
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    this doesn't answer your question, but I recall from grad school long ago, that sometimes in differential geometry $C^2$ is as good as $C^\infty$ (if my recollection is just plain wrong, I can delete this comment). – Stefan Smith Oct 23 '13 at 23:43
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    One important feature of the space $C^\infty$ is that the derivatives are endomorphisms $\frac{\partial}{\partial x_i} : C^\infty \to C^\infty$, whereas otherwise you only have $\frac{\partial}{\partial x_i} : C^k \to C^{k-1}$. This is also (one of) the reason(s) why the "derivation" approach to tangent vectors does not "work" for $C^k$ manifolds (see the answer of @KevinCarlson). – PhoemueX Sep 03 '14 at 16:10
  • @PhoemueX where can I find this proof? I want to prove that on Ck manifold it can be defined only C(k-1) vector fields. – Danilo Lombardo Mar 12 '24 at 15:29

2 Answers2

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I found something interesting! Let $M$ be a $C^k$ manifold, $1\leq k\leq \infty$ and $\mathcal{O}(M)_P$ its local ring of germs of $C^k$ functions at some point $P\in M$. One definition of tangent vector is a derivation on $\mathcal{O}(M)_P$, that is, a linear map $\alpha:\mathcal{O}(M)_P\to \Bbb{R}$ such that $\alpha(fg)=\alpha fg(P)+f(P)\alpha(g)$. Then it becomes necessary to prove that the vector space $A_p$ of derivations of $\mathcal{O}(M)_P$ has the same dimension as $M$.

In case $k=\infty,$ one can do this (following Warner, Foundations of Differentiable Manifold and Lie Groups) by indicating an isomorphism between $A_p$ and $(\mathfrak{m}_P/\mathfrak{m}^2_P)^*$, where $\mathfrak{m}_P$ is the maximal ideal of $\mathcal{O}(M)_P,$ i.e. the space of germs of functions vanishing at $P$. This isomorphism exists whether $k$ is finite or infinite.

But then we have to show $\mathfrak{m}_P/\mathfrak{m}^2_P$ is $n$-dimensional, and for $k$ finite this is false! The proof for $k=\infty$ is an extension of the simple calculus exercise that shows every smooth map $f:\Bbb{R}\to\Bbb{R}$ with $f(0)=0$ is divisible by $x$, i.e. the coordinate function $x$ generates the ideal $\mathfrak{m}_0(\Bbb{R})$. One does this by observing that $f=xg$ for $g(x)=\int_0^1 f'(tx)dt$. This is the point where we need infinite $k$: for $k<\infty,$ $g$ need not be a $C^k$ function!

It actually turns out that for $C^k$ manifolds with finite $k$, the space of derivations $A_P$ is infinite-dimensional, and so can't serve as a model of tangent space at all. The relevant issue is exactly that the product of two $C^k$ functions vanishing at $P$ is in $C^{k+1}$ near $P$; this permits derivations which don't vanish on every $C^k$ function with zero partial derivatives.

Kevin Carlson
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    I learned this from MO some months ago and was just about to post it! – Ryan Reich Oct 24 '13 at 15:20
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    Nice answer! Consequently for $k<\infty$ the tangent space is not modeled by locally ringed space morphisms from the disembodied tangent vector $\operatorname{Spec}\mathbb R[\varepsilon]$. – Arrow Oct 13 '19 at 10:01
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One very useful fact is that if $G$ is an open subset of $\mathbb{R}^d$, then $\mathcal{C}^\infty_c(G)$ is dense in $\mathcal{L}^p(G)$ if $1\le p < \infty$.

ncmathsadist
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    Right, I remember this now...but is there any reason this is a better theorem than "$\mathcal{C}^k_c(G)$ is dense in $\mathcal{L}^p(G)$ for all integers $k$"? – Kevin Carlson Oct 23 '13 at 23:26
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    You do not have to worry about counting derivatives. These functions are also dense in critical Sobolev spaces that appear in the study of PDEs. – ncmathsadist Oct 23 '13 at 23:30
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    It says that the intersecttion of the $\mathcal{C}^k$ is dense. This is a far stronger statement than saying each $\mathcal{C}^k$ is dense. – ncmathsadist Oct 23 '13 at 23:31
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    A strong statement indeed, but it doesn't seem to answer the question to me - this shows that moving from $C^k$ to $C^\infty$ doesn't lose the density result, but it is not motivation to make the move in the first place. – Anthony Carapetis Oct 24 '13 at 00:27
  • Exactly as Anthony Carapetis said-certainly it's stronger, but what are the fruits of that strength, if any? – Kevin Carlson Oct 24 '13 at 00:43